# need help with limit, just a refresher

• Sep 22nd 2009, 11:58 AM
ffxwolf
need help with limit, just a refresher
Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3
• Sep 22nd 2009, 12:28 PM
redsoxfan325
Quote:

Originally Posted by ffxwolf
Prove that the stated limit is correct by finding a number d (delta) in terms of E (is an element of) such that
|f(x) - L| < E whenever |x - c|< d:

lim (4x - 1) = 11
x -> 3

You want to find $\delta$ such that $|x-3|<\delta \implies |(4x-1)-11|<\epsilon$.

Choose $\delta=\frac{\epsilon}{4}$.

Thus when $|x-3|<\delta$, we have that $|(4x-1)-11|=|4x-12|=4|x-3|<4\delta=\frac{4\epsilon}{4}=\epsilon$
• Sep 22nd 2009, 12:43 PM
ffxwolf
Thank you for your response, I just don't understand why delta = epsilon/4
• Sep 22nd 2009, 12:53 PM
redsoxfan325
Quote:

Originally Posted by ffxwolf
Thank you for your response, I just don't understand why delta = epsilon/4

If you don't know what $\delta$ is, you can solve for it.

Whatever $\delta$ is you know that $|x-3|<\delta$.

So, work with what you have. You ultimately want $|(4x-1)-11|<\epsilon$.

So, similar to before,

$|(4x-1)-11|=|4x-12|=4|x-3|<4\delta$.

Now you want $4\delta$ to be equal to $\epsilon$, so you just solve for it.

$4\delta=\epsilon\implies \delta=\frac{\epsilon}{4}$
• Sep 22nd 2009, 12:58 PM
ffxwolf
Okay, I see it now. Thanks (Nod)